Let $(\Omega,\Sigma)$ be a measurable space. An atom of $\Sigma$ is a set $B\in\Sigma$ such that for all $A\subseteq B$ either $A=\emptyset$ or $B=A$. A measurable space is atomic if every element lies in some atom. The $\sigma$-algebra $\Sigma$ is countably generated if there is a countable family of measurable sets such that $\Sigma$ is the smallest $\sigma$-algebra containing all of them. For example $(\mathbb{R},\mathcal{B})$ is countably generated since $\mathcal{B}$ is generated by the open intervals with rational endpoints. The atoms of $\mathcal{B}$ are the singletons.
Proposition: If $\Sigma$ is countably generated, then $(X,\Sigma)$ is atomic.
Proof: If there is a countable family generating $\Sigma$, there is also a countable family closed under complementation that generates $\Sigma$. If $\mathcal{C}$ is such a family, we get all atoms of $\Sigma$ as the intersection of all elements of $\mathcal{C}$ that contain a given point.
Now if $(\Omega,\Sigma,\mu)$ is a probability space, we call $B\in\Sigma$ a $\mu$-atom if $\mu(B)>0$ and for all $A\in\Sigma$ such that $A\subseteq B$, either $\mu(A)=0$ or $\mu(A)=\mu(B)$. The probability space is atomless if it contains no $\mu$-atom.
Lemma: If $(\Omega,\Sigma,\mu)$ is a probability space such that $\Sigma$ is countably generated and $\mu$ takes on only the values $0$ and $1$, then there exists an atom $A\in\Sigma$ such that $\mu(A)=1$.
Proof: Let $\mathcal{C}$ be a countable family closed under complementation that generates $\Sigma$. For each element of $\mathcal{C}$, either itself or its complement has probability one $1$. The intersection of all elements in $\Sigma$ with probability $1$ is an atom with probability $1$.
Proposition: If $(\Omega,\Sigma,\mu)$ is a probability space with $\Sigma$ countably generated, then it is atomless if and only if every atom in $\Sigma$ has probability $0$.
Proof: Clearly, in an atomless probability space, every atom must have probability $0$. Supppose now that $A$ is a $\mu$-atom. Let $A\cap\Sigma=\{A\cap S:S\in\Sigma\}$ be the trace $\sigma$-algebra. It is countably generated too. Then $(A,A\cap\Sigma,1/\mu(A)\cdot\mu)$ is a probability space such that the probability takes on only the values $0$ and $1$. So by the lemma, there is an atom $B$ such that $1/\mu(A)⋅\mu(B)=1$. But $B$ is also an atom of $\Sigma$ and $\mu(B)>0$.
So it follows that a probability measure on $(\mathbb{R},\mathcal{B})$ is atomless if and only if it puts probability $0$ on all singletons, which justifies the definition in the book of Kai Lai Chung.
Finally, an example of a probability space in which each atom has probability $0$ but such that the space is not atomless. Let $\Omega$ be any uncountable set, let $\Sigma$ consists of those subsets of $\Omega$ that are either countable or have an uncountable complement. Let $\mu(A)=0$ if $A$ is countable and $\mu(A)=1$ if its complement is countable. Every set with countable complement is an $\mu$-atom, but the atoms of $\Sigma$ are the singletons which all have probability $0$. Note that $\Sigma$ is not countably generated.
Best Answer
No. Trivial example: suppose $\mu$ is Lebesgue measure and $\nu(S)=\begin{cases}0&|S|=\omega\\\infty&|S|\geq\omega_1\end{cases}$. Then $\mu\times\nu=\nu$ (morally — the product is defined by the same formula), which is atomic.
However, if one requires $\mu$ and $\nu$ to be ($\sigma$-)finite measures, then the product is indeed atomless.
Let $\mu$ be atomless and $\nu$ arbitrary, and write $\lambda=\mu\times\nu$. We will show $\lambda$ lacks atoms, by transfinite induction on the rank over the generators.1 For this induction, it is useful to adopt a slightly stronger-appearing definition of atomlessness:
Note that either definition of non-atomicity is preserved under finite unions/intersections/complements, so the only challenge is countable ones. Proving that the definitions are equivalent is a bit tricky, but luckily Wikipedia's page on atomic measures includes a sketch, so I'll defer to them and just jump straight to the induction.
For a base case, suppose $A=B\times C$ for some $B$, $C$. If $A$ is null, the claim is trivial, so suppose $0<\lambda(A)=\mu(B)\nu(C)$, i.e. $\mu(B)>0$. Since $\mu$ is atomless, there exists $D\subseteq B$ with relative measure at most between $\delta$ and $\epsilon$; taking Cartesian products preserves relative measure, so $D\times C$ witnesses that $A$ is not atomic. Thus no rank-0 sets are atomic.
If $\alpha$ is a limit ordinal, then any set of rank at most $\alpha$ in fact has lower rank, so it remains to show the successor ordinal case.
Suppose all rank-$\alpha$ sets are not atoms and $A$ is rank-$(\alpha+1)$ and has positive measure. Either $A=\bigcup_{j\in\omega}{A_j}$ or $A=\bigcap_{j\in\omega}{A_j}$, where each $A_j$ has rank at most $\alpha$. Since our definition of atomic is symmetric under taking complements, it suffices to assume the former. Moreover, we can disjointify the $\{A_j\}_j$: let $\tilde{A}_0=A_0$ and $\tilde{A}_J=\left(\bigcup_{j=0}^J{A_j}\right)\setminus\tilde{A}_{J-1}$ for $J>1$. This doesn't preserve rank, but it does preserve atomicity (see above). By hypothesis, each $A_j$ is not atomic; we now take them disjoint, so that $\lambda(A)=\sum_{j\in\omega}{\lambda(A_j)}$.
Fix $\delta$ and $\epsilon$: for each $j$, there exists $B_j\subseteq A_j$ such that $\delta\lambda(A_j)\leq\lambda(B_j)\leq\epsilon\lambda(A_j)$. By disjointness, taking a union in $j$ adds measures, so that $\delta\lambda(A)\leq\lambda\left(\bigcup_{j\in\omega}{B_j}\right)\leq\epsilon\lambda(A)$, as desired.
1 A quick reminder of rank for $\sigma$-algebras, where I adopt the slightly nonstandard convention that taking complements does not increase rank: let $\mathcal{F}$ be a $\sigma$-algebra generated by $\mathcal{G}$. Define the (transfinite) sequence \begin{align*} \mathcal{F}_0&=\mathcal{G}\cup\{G^{\mathsf{c}}:G\in\mathcal{G}\} \\ \mathcal{F}_{\alpha+1}&=\mathcal{F}_{\alpha}\cup\left\{\bigcup_{j\in\omega}{F_j}:\forall j\,F_j\in\mathcal{F}_{\alpha}\right\}\cup\left\{\bigcap_{j\in\omega}{F_j}:\forall j\,F_j\in\mathcal{F}_{\alpha}\right\} \\ \mathcal{F}_{\alpha}&=\bigcup_{\beta<\alpha}{F_{\beta}}\quad\quad\alpha\textrm{ limit ordinal} \end{align*} It is a fact that $\mathcal{F}=\mathcal{F}_{\omega_1}$ (exercise!).
If $F\in\mathcal{F}_{\alpha}$, then we say $F$ has rank (over $\mathcal{G}$) at most $\alpha$; the rank of $F$ is the minimal such $\alpha$.